Numbers at the bitter end

A frictionless box slides at 10 m/s, horizontally. How tall of a ramp can it climb before it stops and slides back?

The box has kinetic energy 1/2·10·10². After it goes up the ramp, that'll all be converted to potential energy, 10·9.8·h. Set these equal and solve. 1/2·10·10² = 10·9.8·h, so h = (1/2·10·10²) / (10·9.8).

If, instead, you find that the box goes at 6 m/s, how would that change the solution. This replaces a "10" with a "6", but it's not obvious which, as you see three "10"s in the formula.

Maybe you tracked units. The box had kinetic energy 1/2·(10 kg)·(10 m/s)². Potential energy would be (10 kg)·(9.8 m/s²)·h. Then h = (1/2·(10 kg)·(10 m/s)²) / ((10 kg)·(10 m/s²)). In that, you can see the units "m/s" and match it to the "m/s" of a new speed. That tells you which "10"s to replace.

If the new speed is given as 15 mph, you also must notice that mph is commensurable with m/s. If the mass of the box was given from its density and edge-length, and speed from distance moved in a given time, you would have the same units (m) show up twice. Change either the size of the box or the distance it went, and you'll need much context to know which numbers to change.

Or you could avoid risking all these confusions.

Write it all in algebraic form. At the bitter end, put in numbers. — Professor J.F.D.

Variable names encode both the units and, vaguely, the meaning of any value, and tend to be shorter than either. Equations that you apply tend to be stated in terms of named variables. So you can bring in another equation with fewer substitutions each time, often zero, to advance the solution. Your solution needs numbers and units only when it must become a number, at the end, substituted in one step.